Particular solutions of 3 linear PDEs with nonstandart right-hand side

Dear **MHF **members,

I have the following 3 PDEs, whose solutions I know but I am looking for nice explanary solutions. Please help me.

- $\displaystyle (D_{x}^{2}+D_{x}D_{y}-2D_{y}^{2})z=8\big(\sec(2x+y)\big)^{2}\tan(2x+y)$ [Solution: $\displaystyle z_{p}=\tan(2x+y)$]
- $\displaystyle (D_{x}^{2}+D_{x}D_{y}-D_{x}-D_{y})z=1/y^{2}$ [Solution: $\displaystyle z_{p}=1/y$]
- $\displaystyle (D_{x}^{2}-D_{y}^{2}-D_{x}+D_{y})z=(x+1)/x^{2}$ [Solution: $\displaystyle z_{p}=-\ln(x)$]

Thanks.

**bkarpuz**

Re: Particular solutions of 3 linear PDEs with nonstandart right-hand side

What do you mean by "nice explanary solutions"? Do you mean how they came up with $\displaystyle z_p$? If so, I think I can explain this.

Re: Particular solutions of 3 linear PDEs with nonstandart right-hand side

Quote:

Originally Posted by

**Danny** What do you mean by "nice explanary solutions"? Do you mean how they came up with $\displaystyle z_p$? If so, I think I can explain this.

Exactly!

Re: Particular solutions of 3 linear PDEs with nonstandart right-hand side

The second and third are rather straight forward. For the second, the RHS is only a function of y so sek a particular solution in the form $\displaystyle z_p = f(y). $ You'll get an ODE for $\displaystyle f$ (and it's linear!) Similarly for the third. For the first, you'll notice that the RHS is a function of $\displaystyle 2x+y$ so seek $\displaystyle z_p = f(2x+y). $ Since the LHS of the PDEs are independent of both independent variables, this method works.

Hope it helps.

Re: Particular solutions of 3 linear PDEs with nonstandart right-hand side

Quote:

Originally Posted by

**Danny** The second and third are rather straight forward. For the second, the RHS is only a function of y so sek a particular solution in the form $\displaystyle z_p = f(y). $ You'll get an ODE for $\displaystyle f$ (and it's linear!) Similarly for the third. For the first, you'll notice that the RHS is a function of $\displaystyle 2x+y$ so seek $\displaystyle z_p = f(2x+y). $ Since the LHS of the PDEs are independent of both independent variables, this method works.

Hope it helps.

For the first one, I also noticed that $\displaystyle \frac{\mathrm{d}^{2}\tan(u)}{\mathrm{d}u^{2}}=\big (\sec(u)\big)^{2}\tan(u)$ as the left-hand side is involves only 2nd order derivatives.

Dor the 2nd and the 3rd ones, I rewrote left-hand side as $\displaystyle (D_{x}-1)(D_{x}+D_{y})$ and $\displaystyle D_{x}(D_{x}-1)-D_{y}(D_{y}-1)$ or $\displaystyle (D_{x}+D_{y}-1)(D_{x}-D_{y})$ but I could not see any nice things here...

Is it possible to make change of variables to reduce the equation into first-order equation as in the ordinary case?

Thanks...

Re: Particular solutions of 3 linear PDEs with nonstandart right-hand side

As you've said, for the first, you could made a change of variables where u = 2x + y and pick another independent variable.

For the second and third, since you've noticed that the operators factor, you can introduced a new variable as to make your problem two linear first order PDEs. For example, for the second one, if you let

$\displaystyle w = z_x + z_y$ then $\displaystyle w_x - w = \frac{1}{y^2}$

Solve the second, then use this in the first.